2019 ASRJC Prelim H2 Chem P2 ANS

ASRJC JC2 PRELIMS 2019 9729/02/H2 ANDERSON SERANGOON JUNIOR COLLEGE H2 Chemistry 9729 2019 JC2 Prelim Exam Paper 2 Solutions 1 (a) Reducing power of Group 2 elements increases [1] down the group. Number of (principal) quantum shell / electronic shell increases and valence electrons are further away and more shielded from the nucleus. Valence electrons are less strongly attracted to the nucleus and smaller amount of energy is needed to remove the valence electron so it is more easily oxidised. [1] (b) (i) Copper can give up both 3d and 4s electrons for delocalisation whereas calcium can only give up 2 electrons . [1] Electrons act as mobile charge carriers [1] to conduct electricity. (ii) Cu has smaller atomic radius and higher atomic mass [1] compared to Ca. Also, Cu has a more closely packed structure due to stronger metallic bonding as compared to Ca. [1] Hence, Cu has a much higher density than Ca. (c) (i) A Each orbital can hold a maximum of 2 electrons of opposite spin. Diagram B has a pair of electrons having the same spin. [1] with explanation (ii) 3𝑑𝑧2 [1] 3𝑑𝑥𝑧 [1] Alternative answers: 3𝑑𝑥2−𝑦2, 3dxy, 3dyz (d) (i) Both silicon and silica have a giant molecular structure. [1] (ii)  From the data booklet: Si–Si BE: 222 kJ mol–1 Si–O BE: 460 kJ mol–1 [1]  Size of valence orbitals: Si > O  Valence orbitals of Si are more diffused than that of O  Overlap of orbitals is less effective in Si–Si than in Si–O  The Si–O bond in silica is stronger than the Si–Si bond in silicon.  Hence, bond energy: Si –Si < Si–O and more energy is required to break the Si–O bond than the Si–Si bond. [1]

ASRJC JC2 PRELIMS 2019 9729/02/H2 2 2 (a) (i) presence of carboxylic acid or –COOH group presence of CH3CO– or methyl ketone absence of chiral carbon any 2 for [2] (ii) C D CH3 C O CH2CH2COOH [1] each (iii) E: (CH3)2CO; F: CH3CH2CHO [1] each (iv) G: CH3CH=CHCH(CH3)2 [1] (b) (i) Br Br Br Br [1] Orange–red / red–brown Br2 decolourises / turns colourless. [1] (ii) NaOH in ethanol, heat (under reflux) [1] (iii) BrBr [1] 3 (a) (i) [1]: diagram As shown on the diagram, in the presence of catalase, at a certain temperature T,  Catalase provides a different pathway of a lower activation energy (Ea’ < Ea).  More molecules have energies greater than or equal to the lowered activation energy Ea’.  This results in an increase in the frequency of effective collisions. [2] Ea Ea’ Number of molecules with energy Energy Represents no. of molecules with energy greater than or equal to Ea (uncatalysed) Represents no. of molecules with energy greater than or equal to Ea’ (catalysed by catalase)

ASRJC JC2 PRELIMS 2019 9729/02/H2 3 (ii) Repeat the experiment described but using varying volumes of H2O2 and topping it up with deionised water to keep the total volume constant at 50 cm 3 and add in 1 cm3 of catalase. [1] Record the time taken for 20 cm3 of O2 to form. [1] As rate is inversely proportional to the time taken, plot a graph of 1/t against [H 2O2] / VH2O2. [1] Or find the product of (VH2O2) nt and determine the value of n when (VH2O2)n t becomes a constant. (iii) [1]: diagram At low [H2O2], rate is directly proportional to [H2O2] (OR 1st order reaction) since active sites are not fully occupied. At high [H2O2], rate becomes constant (OR zero order reaction) since active sites are fully occupied. [1] (b) (i) FeS is precipitated first. (ii) When FeS is precipitates, [S2–] = 3.7 x 10–19 ÷ 0.100 = 3.70 x 10–18 mol dm–3 (iii) When MnS starts to precipitate, [S2–] = 3.7 x 10–13 ÷ 0.100 = 3.70 x 10–12 mol dm–3 [1] [Fe2+] = 3.7 x 10–19 ÷ 3.7 x 10–12 = 1.00 x 10–7 mol dm–3 (iv) % of Fe2+ remaining in solution = 1.00 x 10–7 ÷ 0.1 x 100% = 0.0001% << 1%  separation is effective (v) As the pH increases, [H+] decreases, the positions of equilibrium (1) and (2) shift to the right, resulting in a higher [S2–]. For se lective precipitation, the pH is increased such that the ionic product of FeS > Ksp(FeS) but ionic product of MnS < Ksp(MnS). Only FeS is precipitated. [1] explain relationship between pH and [S2–] using LCP [1] linking [S2–] to ionic product and hence Ksp to show sequence of precipitation (c) (i) 2Cr2O3 + 3O2 + 8OH– → 4CrO42– + 4H2O [1] (ii) From yellow (CrO42–) to orange (Cr2O72–) to green (Cr3+) [1] (iii) step III: acid–base reaction complex ion in solution P: [Cr(OH)6] 3– (accept [Cr(OH)4]– or [Cr(OH)4(H2O)2]–) [1] (iv) Na3Fe(CN)6 [1] [H2O2] rate

ASRJC JC2 PRELIMS 2019 9729/02/H2 4 4 (a) Ethylamine is a stronger base than phenylamine. Ethylamine: Ethyl group is electron–donating and increases the electron density around the N atom. This increases the availability of the lone pair of electrons on N atom to accept a proton. [1] Phenylamine: Lone pair of electrons on N atom in phenylamine is delocalised into the  electron cloud of the benzene ring. This decreases the availability of the lone pair of electrons on N atom to accept a proton. [1] (b) (i) methyl orange The working pH range of methyl orange lies within the range of rapid pH change of the titration curve which lies in the acidic pH region (equivalence pH < 7). [1] (ii) At equivalence point: CH3CH2NH2 + HCl → CH3CH2NH3+Cl– CH3CH2NH3+ + H2O ⇌ CH3CH2NH2 + H3O+ [H3O+] at equivalence point = 10–5.91 mol dm–3 Ka of CH3CH2NH3+ = 10–(14–3.35) = 10–10.65 + + 2 3 3 2 2 3 ++ 3 2 3 3 2 3 [H O ][CH CH NH ] [H O ]=[CH CH NH ] [CH CH NH ] aK  [CH3CH2NH3+] at equivalence point = 5.91 2 -3 10.65 (10 ) = 0.0676 mol dm(10 )   [1] (iii) Let x cm3 be the volume of HCl needed to neutralise ethylamine. n(HCl) required to neutralise CH3CH2NH2 = n(CH3CH2NH3+) = 0.2 0.0002 mol1000 x x [CH3CH2NH3+] at equivalence point = -3 0.0002x = 0.0676 mol dmx + 20 1000 x = 10.2 cm3 [1] (iv) n(CH3CH2NH2) = 0.2 x 10.2/1000 = 0.00204 mol [CH3CH2NH2] = -30.00204 = 0.102 mol dm20 1000 [1] (v) -+ -2 3 2 3 3 2 2 3 2 2 [OH ][CH CH NH ] [OH ]=[CH CH NH ] [CH CH NH ] bK  - -3.35 -3[OH ] = 10 x 0.102 = 0.0067499 mol dm pOH = –log 0.0067499 = 2.17 pH = 14 – 2.17 = 11.8 [1] (vi) n(HCl) unreacted = 25 - 10.2 x 0.2 = 0.00296 mol1000

ASRJC JC2 PRELIMS 2019 9729/02/H2 5 [HCl] after 25.00 cm3 has been added = [H3O+] = -30.00296 = 0.06577 mol dm20 + 25 1000 pH = –log 0.06577 = 1.18 [1] (vii) (c) (i) Initial amount of C6H5NH2 = 200 x 0.15 = 0.03 mol1000 Initial amount of HCl = 100 x 0.2 = 0.02 mol1000 C6H5NH2 + HCl → C6H5NH3+Cl– + H2O Initial mol 0.03 0.02 Change in mol –0.02 –0.02 +0.02 Final mol 0.01 0 0.02 [1]  - -+ -4.636 5 3 6 5 2 0.02[OH ]( )[OH ][C H NH ] 0.3 = = 100.01[C H NH ] ()0.3 bK [OH–] = 1.1721 x 10–5 mol dm–3 pOH = –log 1.1721 x 10–5 = 4.93 pH = 14 – 4.93 = 9.07 [1]. (ii) C6H5NH3+ + OH– → C6H5NH2 + H2O [1] Volume of HCl /cm3 pH 11.8 5.91 10.65 1.19 (25) 10.2 5.1

ASRJC JC2 PRELIMS 2019 9729/02/H2 6 5 (a)